I have this exercise:
H is a complex hilbert space. And T is a compact operator on H. Show that if H is not separable, then 0 is an eigenvalue of T.
Hint: Use lemma 1, and theorem 2.
The lemma it refers to is this:
Lemma 1:
Let Y be a closed linear subspace of a Hilbert space H. Show that if $Y\ne H$, then $Y^\perp\ne\{0\}$.
The theorem it refers to is this:
Theorem 2:
Let X and Y be normed spaces and let T be a linear operator($T\in L(X,Y)$). Then
if T is compact then $\text{Im } T$ and $\overline{\text{Im T}}$ are seperable.
The solution is this:
Since H is not seperable it follows from Theorem 2 that $\overline{\text{Im T}}\ne H$, so Ker T =$\overline{\text{Im T}}^\perp\ne\{0\}$(see Lemma 1). Thus there exists $e \ne0$ such that $Te=0$, that is, e is an eigenvector of T, with eigenvalue 0.
But they use that Ker T =$\overline{\text{Im T}}^\perp$. How do they get this?
That result quoted in the solution is not quite true: what is true is that $\text{Ker } T = (\text{Im } T^*)^{\perp}$. It doesn't matter, though, since $T$ being compact implies that its Hilbert space adjoint $T^*$ is compact as well. Hence, by Theorem 2, $(\text{Im } T^*)^{\perp} \neq \{0\}$, so that there is an $e \neq 0$ in $H$ so that $Te=0$.